Saved Bookmarks
| 1. |
A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically downwards with an initial speed of 10 m/s. If acceleration due to gravity is 10 m//s^(2), the separation between the fragments, 2 seconds after the explosion is |
|
Answer» Solution :Initial relative VELOCITY `u_(rel)=10-(-10)=20m/s` Relative acceleration `a_(rel)=g-g=0` After t = 2 SEC, relative separation `S_(rel)=u_(rel)t+(1)/(2)a_(rel)t^(2)=(20xx2)+0=40 m` |
|